1. The first term.

2. The last term.

3. The number of terms.

4. The equal difference or ratio.

5. The sum of all the terms.

Note. As the last term in a long series of numbers is very tedious to come at, by continual multiplication; therefore, for readier finding it out, there is a series of numbers made use of in Arithmetical Proportion, called indices, beginning with an unit, whose common differenceis one; whatever number of indices you make use of, set as srany numbers (in such Geometrical Proportion as is given in the puestion) under them.

As

1, 2, 3, 4, 5, 6 Indices.

2, 4, 8, 16, 32, 64. Numbers in geometrical proportion. But if the first term in geometrical proportion be different from the ratio. the indices must begin with a cypher

As

0, 1, 2, 3, 4, 5, 6. Indices.

1, 3, 4, 8, 16, 32, 64. Numbers in geometrical proportion. When the indices begin with a cypher, the sum of the indices made choice of must always be one less than the number of terms, given in the question; for 1 in the indices is over the second term, and 2 over the third, &c.

Add any two of the indices together, and that sum will agree with the product of their respective terms,

4 X 32

As in the first table of indices 2 + 5 =
Geometrical proportion
Then the second

128

2 + 4 = 6

4 X 16 = 64

In any geometrical progression proceeding from unity, the ratio being known, to find any remote term, without producing all the intermediate terms.

RULE. Find what figure of the indices added together would give the exponent of the term wanted; then multiply the numbers standing under such exponent into each other, and it will give the term required.

NOTE. When the exponent 1 stands over the second term, the number of exponents must be 1 less than the number of terms.

EXAMPLES.

1. A man agrees for 12 peaches, to pay only the price of the last, reckoning a farthing for the first, and a halfpenny for the second, &c. doubling the price to the last, what must he give for them?

0, 1, 2, 3, 4, Exponents.

1, 2, 4. 8, 16. No of terms.

Ans. £...2...8.

For 4+4+3=1. No. of terms less, 1

4)2048—11 No. of far.

12)512

210)42...8

£2...2...8

2. A country gentleman going to a fair to buy some oxen, meets with a person who had 23; he demanded the price of them, was answered £16 a piece: the gentleman bids him £15 a piece, and he would buy all; the other tells him it could not be taken; but if he would give what the last ox would come to, at a farthing for the first and doubling it to the last, he should have all. What was the price of the oxen? Ans. £4369...1...4

In any geometrical progression not proceeding from unity, the ratio being given, to find any remote term, without producing all the intermediate terms.

RULE. Proceed as in the last, only observe that every product must be divided by the first term.

3. A sum of money is to be divided among 8 persons, the first to have £20, the second £60, and so in triple proportion; what will the last have? Ans. £43740.

3+3+1=7, one less than the number of terms. 4. A gentleman dying left nine sons, to whom and to his executors, he bequeathed his estate in manner following: To his executors £50. his youngest son was to have as much more as the executors, and each son to exceed the next younger by as much more; what was the eldest son's portion? Ans. ££5600.

The first term, ratio, and number of terms given, to find the sum of all the terms.

RULE. Find the last term as before, then subtract the first from it, and divide the remainder by the ratio, less 1; to the quotient of which add the greater, gives the sum required.

EXAMPLES.

5. A servant skilled in numbers agreed with a gentleman to serve him twelve months, provided he would give him a farthing for his first month's service, a penny for the second, 4d. for the third, &c. what did his wages amount to? Ans. £58258.....5. 256X256 65536, then 65536 × 644194304

0, 1, 2, 3, 4, 1, 4, 16, 64, 256.

4194304-1
4-1

++4+3=11. No. of terms less 1,

1398191, then

1398101+4194304&592405 farthings.

6. A man bought a horse, and by agreement was to give a farthing for the first nail, three for the second, &c. there were four shoes, and in each shoe 8 nails; what was the worth of the horse? Ans. £965114681693...13...4.

7. A certain person married his daughter on New-year's day, and gave her husband 1s. towards her portion, promising to double it on the first day of every month for 1 year; what was her portion? Ans. £204...15.

8. A laceman, well versed in numbers, agreed with a gentleman to sell him 22 yards of rich gold brocaded lace, for 2 pins the first yard, 6 pins the second, &c. in treble proportion; I desire to know what he sold the lace for, if the pins were valued at 100 for a farthing; also what the laceman got or lost by the sale thereof, supposing the lace stood him in £7. per yard? Ans. The lace sold for £326886...0...9. Guin £326732...0...9.

PERMUTATION

S the changing or varying the order of thing.

RULE. Multiply all the given terms one into another, and the last product will be the number of changes required.

EXAMPLES.

1. How many changes may be rung upon 12 bells; and how long would they be ringing but once over, supposing 40 change. might be rung in 1 minute, and the year to contain 365 days, hours?

1 x 2 x 3 x 4 × 5 × 6 × 7 × 8 × 9 × 10×11×12=479001600 changes, which 10=47900160 minutes; and if reduced, is 91 years, 3 weeks, 5 days, 6 hours. 2. A young scholar coming into town for the convenience of a good library, demands of a gentleman with whom he lodged, what his diet would cost for a year, who told him £10. but the scholar not being certain what time he should stay, asked him what he must give him for so long as he should place his family (consisting of 6 persons besides himself) in different positions, every day at dinner; the gentleman thinking it would not be long, tells him £5, to which the scholar agreos. What time did the scholar stay with the gentleman? Ans. 5040 days.

(T.S.) In an establishment of 100 pupils, I desire to know how many different positions you can place them in at dinner, and what would be the amount of board and education for one young gentleman, provided he remained at school during the above changes, terms being 40 guineas per annum, and what would the master gain by his establishment, provided he had the above number of pupils all the time of the changes; his profit being 10 per cent. and what would the pupils receive in pocket money, half of them to have 2d., 3d., 4d. and the rest Gd. per week?

THE

TUTOR'S ASSISTANT.

PART II.

VULGAR FRACTIONS.

A FRACTION is a part or parts of an unit, and written with

two figures, with a line between them as, }, }, &c.

The figure above the line is called the numerator, and the under one the denominator; which shews how many parts the unit is divided into; and the numerator shews how many of those parts are meant by the fraction.

There are four sorts of vulgar fractions: proper, improper, compound, and mixed, viz.

1. A PROPER FRACTION is when the numerator is less than the denominator, as,,,,, &c.

2 3

101

2. AN IMPROPER FRACTION is when the numerator is equal to, or greater than the denominator, as,,,197, &c.

3. A COMPOUND FRACTION is the fraction of a fraction, and known by the word of, as, of, of 3, of I, of ů, &c.

4. A MIXED NUMBER OR FRACTION is composed of a whole number and fraction, 83, 174, 87, &c.

1.

71 979

REDUCTION OF VULGAR FRACTIONS.

To reduce fractions to a common denominator,

RULE. Multiply each numerator into all the denominators, except its own, for a new numerator; and all the denominators for a common denominator. Or,

2. Multiply the common denominator by the several given numerators separately, and divide their product by the several denominators, the quotient will be the new numerators.

EXAMPLES.

16

1. Reduce and to a common denominator. Facit, and 18. 1st. num.

2d. num.

16

2X7 14 4X4 16, then 4X7 28 den., and 18. 2. Reduce,, and to a common denominator.

Facit,,.

3. Reduce,,, and, to a common denominator.

Facit 2940

2740 2016 3360 3366

4. Reduce,,,, to a common denominator.

Facst

2380

33609 3360°

1008 840
16809 1780

240 840 57809 7650°

5. Reduce, 1, 3, and to a common denominator. Facit 0840, 340 672 560 860 105

6. Reduce,,, and to a common denominator. Facit 2160 21609 2160 213

720 1200 540 1295

2. To reduce a vulgar fraction to its lowest terms. RULE. Find a common measure by dividing the lower term by the upper, and that divisor by the remainder following, till nothing remain; the last divisor is the common measure; then divide both parts of the fraction by the common measure, and the quotient will give the fraction required.

Note. If the common measure happens to be one, the fraction is already in its lowest term; and when a fraction hath cyphers at the right hund, it may be abbreviated by cutting them off, as 30.

to its lowest terms.

to its lowest terms.

12. Reduce 5184 to its lowest terms.

6972

3. To reduce a mixed number to an improper fraction. RULE. Multiply the whole number by the denominator of the fraction, and to the product add the numerator for a new numerator, which place over the denominator.

Note. To express whole numbers fraction-ways, set 1 for the denominator given.

EXAMPLES.

13. Reduce 183 to an improper fraction.

18x7+3=129 new numerator,=120. 14. Reduce 5613 to an improper fraction. 15. Reduce 183 to an improper fraction. 16. Reduce 13 to an improper fraction. 17. Reduce 27 to an improper fraction.

18. Reduce 514, to an improper fraction..

16.

4. To reduce an improper fraction to its proper terms.. RULE. Divide the upper term by the lower.

· す

Facit 183.

Facit 56 Facit 183: